I am concerned with the asymptotical behavior of integrals like this for large $n$
$$\frac{1}{n!}\intop_{\Omega}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}dx_{j},$$
where $\Omega$ is one of these infinite rectangular regions in $\mathbb{R}^n$:
$$\Omega_1^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2<\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$
$$\Omega_2^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$
$$\ldots$$
$$\Omega_n^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3>\lambda\ldots,\,x_n>\lambda\},$$
and $\lambda$ is a large positive number.
Some information about such integrals has already been obtained in https://mathoverflow.net/a/200622/22773

Originally, I am trying to estimate a polynomial sum of integrals over $\Omega_k$. Namely, let $a$ be any number such that $a>1$. For convenience, define

where $f(\lambda)=o(1)$ for large $\lambda$, then the previous statement follows immediately (with, say, $C=2a$).

Edit:
Another important fact that I remembered is that $I_k^{(n)}(\lambda)$ is, up to a constant, the probability that a $n\times n$ random matrix from the Gaussian Unitary Ensemble has exactly $k$ eigenvalues on $(\lambda,+\infty)$. When $n\rightarrow\infty$, the analogue of this is also known as $E(k,\lambda)$ and the asymptotics of these functions as $\lambda\rightarrow -\infty$ have been examined in Tracy&Widom "Level-Spacing Distributions and the Airy Kernel".
Let's call $J=(\lambda,\infty)$ and define $E_k^{(n)}(\lambda)$ as the probability of a random $n\times n$ matrix in the GUE having $k$ eigenvalues in $J$. Then the following expression is known:

So, $R^{(n)}$ is a very common object when studying the Gaussian Unitary Ensemble, but typically one is interested in its values or derivatives at $z=-1$. In my problem, however, values at arbitrary $z$ come up.

It would probably suffice to prove that $R^{(n)}(z,\lambda),\, n=1,2,3,\ldots$ is a convergent sequence of analytic functions in $z$ and it converges uniformly for all large $\lambda$.

It is also worth noting that $R^{(n)}(z,\lambda)$ is the generating function of integrals over $J^k \times \mathbb{R}^{n-k}$ (they are, in fact, integrals over $J^k$ of the $k$-point correlation functions):

2 Answers
2

Calculations for $\boldsymbol n$ up to $\boldsymbol 6$ indicate that, for $1\leq k\leq n$, $0\leq r\leq k-1$, there are polynomials $P_{n,k,r}\in\mathbb Q[\lambda]$ of certain degrees $d_{n,k-r}$ - which are even if $k-r$ is even and odd if $k-r$ is odd - and constants $Q_{n,k}\in\mathbb Q$ such that the following holds true:
$$
\boxed{\sum_{l=k}^n \binom{n-k}{l-k}\,I_l^{(n)}(\lambda) =
\sum_{r=0}^{k-1}e^{-\left(k-r\right)\lambda^2}\pi^{\frac{n-k+r}2}
P_{n,k,r}(\lambda)\,E^r(\lambda) +
Q_{n,k}\,\pi^{\frac{n}2}E^k(\lambda)}
$$
(For the left-hand side: The integral is extended over the region $\lambda<x_i<\infty$ for $1\leq i\leq k$ and $-\infty<x_i<\infty$ for $k+1\leq i\leq n$.) Here, $E(\lambda)=\frac2{\sqrt{\pi}}\int_\lambda^\infty e^{-y^2}dy$ is the
complementary error function. Notice that $E(\lambda)=\frac{e^{-\lambda^2}}{\sqrt{\pi}\,\lambda}\left(1+O\left(\frac1{\lambda^2}\right)\right)$ as
$\lambda\to\infty$. In particular, one then has
$$
I_k^{(n)}(\lambda) \sim c_{n,k}\,\pi^{\frac{n-k}2}\,e^{-k\lambda^2}
\,\lambda^{N_{n,k}} \tag{1}
$$
as $\lambda\to\infty$ for some constant $c_{n,k}\in\mathbb Q$, where $N_{n,k}= \max
\{d_{n,k-r}-r\,\mid\,0\leq r\leq k-1\}$. For $k=1$, this was already observed in this previous post.

Remark. The $P_{n,k,r}$ appear to depend only on $(n,k-r)$, up to a multiplicative
constant.

$\begingroup$Although this is interesting, I am constantly returning to the thought that the actual solution might be much simpler. I added some information to the post which might help in searching for a proof. The quantity that you calculated here is $\rho_l$ in my post. It is possible that some estimates for $R^{(n)}(z,\lambda)$ are already known, since it is such a common object in random matrix theory, but I haven't been able to find anything yet.$\endgroup$
– level1807Mar 23 '15 at 10:37

$\begingroup$Your additional info is certainly helpful. In order to get what is calculated above you need to replace $\chi_{J^n}$ with $\chi_{J^k\times{\mathbb R}^{n-k}}$.$\endgroup$
– ifwMar 23 '15 at 12:38

... I am constantly returning to the thought that the actual solution might be much simpler.

I do not think that you are right. To obtain estimates for fixed $n$ is not hard, the big issue is uniformity with respect to $n$. Of course, it is possible that the specific context allows to play some algebraic tricks which facilitate the computations. But in the end it comes down to getting these integrals estimated, and I am convinced that any argument that does not take into account the fine structure of the problem is doomed to fail.

Here is a heuristic argument to support my point of view: You are seeking to prove an estimate like
$$
I_{k+1}^{(n)}(\lambda) \leq \frac{f(\lambda)}{n}\,I_k^{(n)}(\lambda)
$$
with $f(\lambda)= o(1)$ as $\lambda\to\infty$. Inserting the asymptotics from (1), so neglecting remainder terms, one gets
$$
\frac{c_{n,k+1}}{c_{n,k}}\,e^{-\lambda^2}\lambda^{N_{n,k+1}-N_{n,k}}
\leq \frac{C\, f(\lambda)}{n}.
$$
The calculations performed before suggest that $d_{n,1}=2n-3$, $d_{n,2}\geq 2n-2$, and $d_{n,k}=d_{n,2}-k+2$ for $k\geq2$. If so, then one has $N_{n,1}=2n-3$ and $N_{n,k}=d_{n,2}-k+2$ for $k\geq2$.

$\begingroup$Yes, I noticed that too. Combinatorics in expanding the polynomial is pretty awful. I am also hoping that the traditional GUE theory might provide us with some magical properties of $R^{(n)}$. Specialists might know.$\endgroup$
– level1807Mar 23 '15 at 20:22